Topic: Internal resistance, emf, cells in series and parallel· Current Electricity

The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω as shown in the figure is :

The terminal voltage of the battery, whose emf is 10 \(V\) and internal resistance 1 \(\Omega\), when connected through an external resistance of 4 \(\Omega\) as shown in the figure is :

circuit: 4 ohm external resistor with 10 V battery and 1 ohm internal resistance

1. 10 \(V\)

2. 4 \(V\)

3. 6 \(V\)

4. 8 \(V\)

Doubtly Asked 30 Oct 2026 NEET UG 2024 Practice test neet Save WhatsApp Report

1 answer

Best answer

Answer: 4. 8 \(V\)

Key idea: Terminal voltage \(V=\varepsilon-Ir\).

Current \(I=\frac{\varepsilon}{R+r}=\frac{10}{4+1}=2\) A.

\(V=10-2\times1=8\) V (equivalently \(IR=2\times4=8\) V).

Option 4.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

The terminal voltage of the battery, whose emf is 10 V and internal resistance 1 Ω, when connected through an external resistance of 4 Ω as shown in the figure is :

1 answer

Your answer